2015 HUAS Summer Trainning #5~A

Description

Given a sequence a[1],a[2],a[3]......a[n], your job is to calculate the max sum of a sub-sequence. For example, given (6,-1,5,4,-7), the max sum in this sequence is 6 + (-1) + 5 + 4 = 14. 

 

Input

The first line of the input contains an integer T(1<=T<=20) which means the number of test cases. Then T lines follow, each line starts with a number N(1<=N<=100000), then N integers followed(all the integers are between -1000 and 1000). 

 

Output

For each test case, you should output two lines. The first line is "Case #:", # means the number of the test case. The second line contains three integers, the Max Sum in the sequence, the start position of the sub-sequence, the end position of the sub-sequence. If there are more than one result, output the first one. Output a blank line between two cases. 

 

Sample Input

2

5 6 -1 5 4 -7

7 0 6 -1 1 -6 7 -5

 

Sample Output

Case 1:

14 1 4

 

Case 2:

7 1 6

解题思路：这个题目的意思是求一组数的最大子段和，最大子序列是要找出由数组成的一维数组中和最大的连续子序列。  找最大子序列的方法很简单，只要前i项的和还没有小于0那么子序列就一直向后扩展，否则丢弃之前的子序列开始新的子序列，同时我们要记下各个子序列的和，最后找到和最大的子序列。如果枚举来求解显然是行不通的，他的时间复杂度约为O(n^3),改用递归也会超时，所以最好利用动态规划和分治法相结合（时间复杂度为O(n)）。每一次求解时还要注意子序列的起始位置和重点位置。

程序代码：

#include<stdio.h> 
int a[100005],str[100005],start[100005];
int main()
{ 
 int t,n,i,num=1,end,max,k; 
 scanf("%d",&t); 
 while(t--)  
 {
  scanf("%d",&n);  
  for(i=1;i<=n;i++)  
  {    
   scanf("%d",&a[i]);          
  }   
  str[1]=a[1];
  start[1]=1;  
  for(i=2;i<=n;i++)  
  {   
   if(str[i-1]>=0)   
   {    
    str[i]=str[i-1]+a[i];    
    start[i]=start[i-1];    
   }   
   else   
   {    
    str[i]=a[i];    
    start[i]=i;   
   }   
  }   
  max=str[1];
  end=1;  
  for(k=2;k<=n;k++)  
  {   
   if(str[k]>max)   
   {    
    max=str[k];    
    end=k;   
   }   
  }   
  printf("Case %d:\n",num);  
  num++;   
  printf("%d %d %d\n",max,start[end],end);   
  if(t)          
   printf("\n");   
 } 
 return 0; 
}